Some conventional voltage comparison circuits need voltage trimming. For example, if a voltage at a first terminal is V1, a voltage at a second terminal is V2, a voltage V12 between the first terminal and the second terminal can be given by: V12=V1−V2. In order to compare the voltage V12 with one or more reference voltages to check the cell condition such as over-voltage or under-voltage, the reference voltages may need to be with respect to a reference level that is equal to V2 (or V1). Besides, the reference level may need to be adjusted according to different requirements, which may require a lot of reference voltage trimming process. In some conventional battery protection controllers for multiples cells in series, all the cell voltages may need to be compared with one or more trimmed reference voltages. Such battery protection controller may require a lot of trimming process, which may increase the cost. Therefore, voltage conversion circuits can be used to shift the cell voltages with respect to ground and then to compare cell voltages with one or more predetermined ground-based reference voltages.
FIG. 1 illustrates a block diagram of a conventional voltage-to-voltage conversion circuit 100. As show in FIG. 1, the conversion circuit 100 includes a plurality of cells 102_1-102_4 having a plurality of cell voltages V102—1, V102—2, V102—3 and V102—4 respectively, a plurality of switches 104_1-104_4 coupled to the cells 102_1-102_4, a plurality of switches 106_1-106_4 coupled to the cells 102_1-102_4, and a voltage level shifter 140 coupled between the switches 104_1-104_4 and the switches 106_1-106_4. The voltage level shifter 140 can output a voltage 130 indicative of the cell voltage V102—1, V102—2, V102—3 or V102—4 of each cell 102_1-102_4 by tuning on each switch of the plurality of switches 104_1-104_4 and each switch of the plurality of switches 106_1-106_4 sequentially. For example, when the switch 104_4 and the switch 106_4 are turned on, the voltage level shifter 140 can output a voltage 130 that is proportional to the cell voltage of the cell 102_4 by applications of an operational amplifier 120 and resistors 112, 114, 116 and 118.
When the switches 104_4 and 106_4 are on, if I102—3 is a current flowing through the cells 102_1-102_3, I106—4 is a current flowing through the switch 106_4, a current I102—4 flowing through the cell 102_4 can be given by:I102—4=I102—3−I106—4.As a result, the current I102—4 flowing through the cell 102_4 is unbalanced with the current I102—3 flowing through the cells 102_1-102_3. Similarly, when sampling the cell voltages of the cells 102_1-102_3, the currents flowing through the cells 102_1-102_4 are also unbalanced.
Furthermore, the conversion circuit 100 can only sample the cell voltage of each cell of the cells 102_1-102_4 sequentially, which may not meet the requirement of sampling all the cells voltage simultaneously in many applications.
FIG. 2 illustrates a block diagram of a conventional voltage-to-current conversion circuit 200. As shown in FIG. 2, the conversion circuit 200 includes a plurality of cells 202_1-202_4, and a plurality of voltage-to-current converters 212_1-212_4 respectively coupled to the plurality of cells 202_1-202_4 in parallel. As shown in FIG. 2, each of the converters 212_2-212_4 includes a corresponding operational amplifier (204_2, 204_3, or 204_4), a corresponding PMOSFET (206_2, 206_3, or 206_4), and a corresponding resister (208_2, 208_3, or 208_4) respectively. The converter 212_1 includes a resistor 208_1. The converters 212_1-212_4 can convert cell voltages of the cells 202_1-202_4 to currents I208—1, I208—2, I208—3 and I208—4 respectively. For example, a current I208—i flowing through the resistor 208_i can be given by:I208—i=V202—i/R208—i,where V202—i is a cell voltage of the cell 202_i and R208—i is a resistance of the resistor 208_i (i=1, 2, 3, 4).
As shown in FIG. 2, I204—2, I204—3 and I204—4 are the operation currents of the operational amplifiers 204_2-204_4 respectively, and I208—1, I208—2, I208—3 and I208—4 are the currents flowing through the resistors 208_1-208_4 respectively. If the voltages V202—1, V202—2, V202—3 and V202—4 are equal to the same voltage V202, and the resistors R208—1, R208—2, R208—3 and R208—4 have the same resistance R208, then the currents I208—1, I208—2, I208—3 and I208—4 are equal to the same current I208 (I208=V202/R208). Besides, the currents I204—2, I204—3 and I204—4 can be equal to the same current I204. As such, the currents I202—1, I202—2, I202—3 and I202—4 respectively flowing through the cells 202_1-202_4 can be given by:
                                          I                          202              ⁢              _              ⁢              4                                =                    ⁢                                    I                              204                ⁢                _                ⁢                4                                      +                          I                              208                ⁢                _                ⁢                4                                      +                          I                              204                ⁢                _                ⁢                3                                      +                          I                              204                ⁢                _                ⁢                2                                                                                              =                        ⁢                                          3                *                                  I                  204                                            +                              I                208                                              ,                                                              I                          202              ⁢              _              ⁢              3                                =                    ⁢                                    I                              202                ⁢                _                ⁢                4                                      +                          I                              208                ⁢                _                ⁢                3                                                                                              =                        ⁢                                          3                *                                  I                  204                                            +                              2                *                                  I                  208                                                              ,                                                              I                          202              ⁢              _              ⁢              2                                =                    ⁢                                    I                              202                ⁢                _                ⁢                3                                      +                          I                              208                ⁢                _                ⁢                2                                                                                              =                        ⁢                                          3                *                                  I                  204                                            +                              3                *                                  I                  208                                                              ,                                          ⁢          and                                                              I                          202              ⁢              _              ⁢              1                                =                    ⁢                                    I                              202                ⁢                _                ⁢                2                                      +                          I                              208                ⁢                _                ⁢                1                                                                                  =                    ⁢                                    3              *                              I                204                                      +                          4              *                                                I                  208                                .                                                        As a result, the currents I202—1, I202—2, I202—3, and I202—4 respectively flowing through the cells 202_1-202_4 are not balanced with each other.